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Farad Capacitor
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 BULLZ AUDIO CAP 4.4 CAR DIGITAL POWER CAPACITOR 4.4F FARAD 4400W 20VDC LOW ESR  US $45.00
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 BULLZ AUDIO CAP 6.6 CAR DIGITAL POWER CAPACITOR 6.6F FARAD 6600W 20VDC LOW ESR US $57.00
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Here are some more information for Farad Capacitor:
After you have discharged the capacitor due to heavy power draw, the capacitor will need to recharge. How does it recharge? Well, the capacitor sits in between your battery and your amplifier. Once the capacitor is discharged, it will attempt to charge itself by drawing power from your battery. Then, when the amp needs power, the capacitor will supply current from the battery as well as any "extra" current it has saved up. Once its discharged, the vicious cycle continues.
Most common car audio capacitors are rated at 1 farad, while I've seen some reach as high as 5, 10, 20, and higher. No matter how high the rating, a capacitor will still hamper performance. This is because capacitors cannot supply the needed current for any extended duration of time. When your amplifier needs current, it will suck the capacitor dry in less than one second, leaving your capacitor to use your battery to charge up again. Once the capacitor is charged, that extra current can then be transferred onto the amp.
See where I'm going now? If a capacitor can only hold a charge for less than one second, while drawing substantial amounts of current from your battery in an attempt to charge back up while delaying power the amp, then what is the point? There is none. Capacitors are known in the car audio community as a marketing ploy to make people think they need something, when in reality, they really don't. Take your money and invest it in a second battery. A capacitor these days runs between $50-$80. Add a few more dollars and you can get a second battery, which will supply your amplifier with much more power than a capacitor ever could, while not straining your electrical system.
Visit http://www.IsYouGeekedUp.com for more car audio information
How To Repair Dell E153Fpc LCD Monitor With Shutdown Problem
Repairing the DELL E153Fpc LCD Monitor with the complaint of display shutdown after few seconds (power LED still on) was not difficult. Normally when this type of problem happens, we would first suspect the inverter board, then the filter capacitors and lastly the main board. A defective in one of the lamps and a faulty TL1451AC PWM IC or inverter ic may caused the display to shutdown as well. If the lamp is defective, it will send a feedback to pin 3 or 14 of TL1451AC IC and caused it to stop producing output signal thus the display just shut off once you switch it on. The power board is using a SG6841D pwm ic and a FQPF7N80 power fet . If this power section blow, it is quite easy to troubleshoot because it has fewer components. You can also use other part numbers to substitute the power fet.
As usual, after opening up the cover, I would first check for any burnt components, loose connector, cracked board, dry joints and etc before doing the voltage test. As for the lamps, I would check it last because lamps were seldom spoilt. Upon scanning for any bad components before starting the voltage test, I found a filter capacitor that had turned into darker color. Using the ESR meter revealed the ESR ohm has shoot up to 12 ohms. This filter capacitor located at C922 with the value of 1000 micro farad 16 volt. If you don’t have the ESR meter you can always use the digital capacitance meter. The value that I got from measuring the bad capacitor was 57 microfarad. If you don’t have ESR meter, make sure you get one-it’s fast and accurate.
Now, the question is why a filter capacitor can cause the display to shutdown? If you carefully see the picture, I’ve drawn the yellow path to show you how the voltage is flowing. By the way, the voltage at the bad filter capacitor line is 12 volt. The voltage passed through a SMD transistor in the start circuit and flow straight to pin 9 (VCC) of TL1451AC IC. In order to let the 12 volt to flow through the transistor and reach the vcc pin of the ic, first, a ‘on’ signal must be sent from the main board to trigger the transistor. If no signal send to the transistors (off signal), both transistors can’t be turned on and no voltage will follow through the ic. Thus if the main board or any of the start circuit components spoilt, there will be power but no display because the TL1451AC would not send signal to the Royer type L, C resonant and boost circuit, so no high voltage produced by the high voltage transformer and this will lead to no display.
In the above case, the LCD Monitor display shutdown was due to the ripple because of the bad filter capacitor. The ripple caused the TL1451AC IC to be unstable and eventually shutdown the output signal. Replacing only the filter capacitor solved the DELL E153Fpc LCD Monitor shutdown problem.
Conclusion, don’t always presumed that LCD Monitor is very tough or difficult to repair. From the above article, you could see that the caused of the shutdown problem was only a bad capacitor. If you fully understood how LCD Monitor works, then chances for you to successfully repair the monitor is very high but again it depends on the availability of spare parts. If you ever come across any LCD Monitor sent to you for repair for the first time, go ahead and start troubleshoot it. Who knows one day LCD Monitor repair could become one of the major source of income to you and your business.
About the Author
Find out more at the links below. You will be very surprised to discover that LCD and Electronic Component Repairs can make thousands of dollars even for a novice. You can find out the repairing secrets.
Find out How to Repair Any LCD Monitor - It is a breeze!
Learn about Secrets of Electronic Repairs.
What is the charge on the 6 micro-farad capacitor?
Obviously when I ask "what is the charge across capacitor with 6 micro-Farads" I do not mean what is the charge on the total capacitance. If you have 20 V running across the circuit in the diagram what is the charge on the 6 micro-Farad capacitor? Is that simple enough or do I need to simplify it more?
http://www.geocities.com/trunks11111/capacitors.jpg
Charge (Q) = Voltage (V) * Capacitance (C)
You have to break down the voltage though, since the full 20V is not across that capacitor.
Combine the 5microF and 6microF capacitors using 1/Ctot = 1/C1 + 1/C2. Then combine them and the 3microF capacitor using simple addition. Then combine that with the 7microF capacitor using 1/Ctot = 1/C1 + 1/C2 again.
Then use Q = VC to get the total charge.
Then calculate the voltage on the 7microF capacitor V = Q/C using that total charge. Subtract that from the total voltage to get the voltage on the rest.
Then use Q = VC with that voltage and the C from the 2 capacitors (5 and 6 microfarads) in series. THAT is your final answer.
Paper supercapacitor could power future paper electronics
(PhysOrg.com) -- All those paper transistors and paper displays that scientists have been designing can now be powered by an onboard power source, thanks to the development of a new paper supercapacitor. Designed by researchers at Stanford University, the paper supercapacitor is made by simply printing carbon nanotubes onto a treated piece of paper. The researchers hope that the integrated ...
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